© The McGraw-Hill Companies, Inc., Chapter 3 The Greedy Method

© The McGraw-Hill Companies, Inc., The greedy method Suppose that a problem can be solved by a sequence of decisions. The greedy method has that each decision is locally optimal. These locally optimal solutions will finally add up to a globally optimal solution. Only a few optimization problems can be solved by the greedy method.

© The McGraw-Hill Companies, Inc., A Simple Example Problem: Pick k numbers out of n numbers such that the sum of these k numbers is the largest. Algorithm: FOR i = 1 to k Pick out the largest number and delete this number from the input. ENDFOR

© The McGraw-Hill Companies, Inc., Shortest Paths on a Special Graph Problem: Find a shortest path from v 0 to v 3. The greedy method can solve this problem. The shortest path: = 7.

© The McGraw-Hill Companies, Inc., Shortest Paths on a Multi-stage Graph Problem: Find a shortest path from v 0 to v 3 in a multi-stage graph. Greedy method: v 0 v 1,2 v 2,1 v 3 = 23 Optimal: v 0 v 1,1 v 2,2 v 3 = 7 The greedy method does not work.

© The McGraw-Hill Companies, Inc., This problem can be solved by the dynamic programming method which will be introduced later.

© The McGraw-Hill Companies, Inc., Minimum Spanning Trees (MST) It may be defined on Euclidean space points or on a graph. G = (V, E): weighted connected undirected graph Spanning tree: T = (V, S), S  E. Minimum spanning tree(MST): a spanning tree with the smallest total weight.

© The McGraw-Hill Companies, Inc., An Example of MST A graph and one of its minimum spanning trees.

© The McGraw-Hill Companies, Inc., Kruskal ’ s Algorithm for Finding MST Input : A weighted, connected and undirected graph G = (V, E ). Output : A minimal spanning tree for G. T : =  While T contains less than n - 1 edges do Begin Choose an edge (v, w) from E of the smallest weight Delete (v, w) from E If (the adding of (v, w) to T does not create a cycle in T) then Add (v, w) to T Else Discard (v, w) End

© The McGraw-Hill Companies, Inc., An example of Kruskal ’ s algorithm

© The McGraw-Hill Companies, Inc., The Details for Constructing an MST How do we check if a cycle is formed when a new edge is added? By the SET and UNION algorithm introduced later. A tree in the forest is used to represent a SET. If (u, v)  E and u, v are in the same set, then the addition of (u, v) will form a cycle. If (u, v)  E and u  S 1, v  S 2, then perform UNION of S 1 and S 2.

© The McGraw-Hill Companies, Inc., A spanning forest If (3,4) is added, since 3 and 4 belong to the same set, a cycle will be formed. If (4,5) is added, since 4 and 5 belong to different sets, a new forest (1,2,3,4,5,6,7,8,9) is created. Thus we are always performing the union of two sets.

© The McGraw-Hill Companies, Inc., Time Complexity of Kruskal ’ s Algorithm Time complexity: O(|E|log|E|) = O(n 2 logn), where n = |V|.

© The McGraw-Hill Companies, Inc., Prim ’ s Algorithm for Finding an MST Step 1: x  V, Let A = {x}, B = V - {x}. Step 2: Select (u, v)  E, u  A, v  B such that (u, v) has the smallest weight between A and B. Step 3: Put (u, v) into the tree. A = A  {v}, B = B - {v} Step 4: If B = , stop; otherwise, go to Step 2. Time complexity : O(n 2 ), n = |V|. (see the example on the next page)

© The McGraw-Hill Companies, Inc., An Example for Prim ’ s Algorithm

© The McGraw-Hill Companies, Inc., The Single-Source Shortest Path Problem Shortest paths from v 0 to all destinations

© The McGraw-Hill Companies, Inc., Dijkstra ’ s Algorithm to Generate Single Source Shortest Paths Input: A directed graph G = (V, E) and a source vertex v 0. For each edge (u, v)  E, there is a nonnegative number c(u, v) associated with it. |V|= n + 1. Output: For each v  V, the length of a shortest path from v 0 to v. S : = {v 0 } For i : = 1 to n do Begin If (v 0, v i )  E then L(v i ) : = c(v 0, v i ) else L(v i ) : =  End (continued on the next page)

© The McGraw-Hill Companies, Inc., Dijkstra ’ s Algorithm to Generate Single Source Shortest Paths (cont ’ d) For i : = 1 to n do Begin Choose u from V - S such that L(u) is the smallest S : = S ∪ {u} (* Put u into S *) For all w in V - S do L(w) : = min(L(w), L(u) + c(u, w)) End

© The McGraw-Hill Companies, Inc., Vertex Shortest distance to v 0 (length) v 1 v 0 v 1 (1) v 2 v 0 v 1 v 2 (1 + 3 = 4) v 3 v 0 v 1 v 3 (1 + 4 = 5) v 4 v 0 v 1 v 2 v 4 ( = 6) v 5 v 0 v 1 v 3 v 5 ( = 8)

© The McGraw-Hill Companies, Inc., Time complexity of Dijkstra ’ s algorithm is O(n 2 ). It is optimal.

© The McGraw-Hill Companies, Inc., Given two sorted lists L 1 and L 2, L 1 = (a 1, a 2,..., a n1 ) and L 2 = (b 1, b 2,..., b n2 ), we can merge L 1 and L 2 into a sorted list by using the following algorithm.

© The McGraw-Hill Companies, Inc., Linear Merge Algorithm Input: Two sorted lists, L 1 = ( a 1, a 2,..., a n1 ) and L 2 = (b 1, b 2,..., b n2 ). Output: A sorted list consisting of elements in L 1 and L 2. Begin i : = 1 j : = 1 do compare a i and b j if a i > b j then output b j and j : = j + 1 else output a i and i : = i + 1 while (i  n1 and j  n2) if i > n1 then output b j, b j+1,..., b n2, else output a i, a i+1,..., a n1. End.

© The McGraw-Hill Companies, Inc., If more than two sorted lists are to be merged, we can still apply the linear merge algorithm, which merges two sorted lists, repeatedly. These merging processes are called 2-way merge because each merging step only merges two sorted lists.

© The McGraw-Hill Companies, Inc., A Greedy Algorithm to Generate an Optimal 2-Way Merge Tree Input: m sorted lists, L i, i = 1, 2,..., m, each L i consisting of n i elements. Output: An optimal 2-way merge tree. Step 1. Generate m trees, where each tree has exactly one node (external node) with weight n i. Step 2. Choose two trees T 1 and T 2 with minimal weights. Step 3. Create a new tree T whose root has T 1 and T 2 as its subtrees and weight are equal to the sum of weights of T 1 and T 2. Step 4. Replace T 1 and T 2 by T. Step 5. If there is only one tree left, stop and return; otherwise, go to Step 2.

© The McGraw-Hill Companies, Inc., An example of 2-way merging Example: 6 sorted lists with lengths 2, 3, 5, 7, 11 and 13.

© The McGraw-Hill Companies, Inc.,  Time complexity of the 2-way merge algorithm is O(n log n).

© The McGraw-Hill Companies, Inc., Huffman codes In telecommunication, how do we represent a set of messages, each with an access frequency, by a sequence of 0 ’ s and 1 ’ s? To minimize the transmission and decoding costs, we may use short strings to represent more frequently used messages. This problem can by solved by using the 2- way merge algorithm.

© The McGraw-Hill Companies, Inc., An example of Huffman algorithm Symbols: A, B, C, D, E, F, G freq. : 2, 3, 5, 8, 13, 15, 18 Huffman codes: A: 10100B: C: 1011 D: 100E: 00 F: 01 G: 11 A Huffman code Tree

© The McGraw-Hill Companies, Inc., The minimal cycle basis problem 3 cycles: A 1 = {ab, bc, ca} A 2 = {ac, cd, da} A 3 = {ab, bc, cd, da} where A 3 = A 1  A 2 (A  B = (A  B)-(A  B)) A 2 = A 1  A 3 A 1 = A 2  A 3 Cycle basis: {A 1, A 2 } or {A 1, A 3 } or {A 2, A 3 }

© The McGraw-Hill Companies, Inc., Def: A cycle basis of a graph is a set of cycles such that every cycle in the graph can be generated by applying  on some cycles of this basis. The weighted cycle basis problem: Given a graph, find a minimal cycle basis of this graph.

© The McGraw-Hill Companies, Inc., The minimal cycle basis is {A 1, A 2 }, where A 1 ={ab, bc, ca} and A 2 = {ac, cd, da}.

© The McGraw-Hill Companies, Inc., A greedy algorithm for finding a minimal cycle basis: Step 1: Determine the size of the minimal cycle basis, denoted as k. Step 2: Find all of the cycles. Sort all cycles by weights. Step 3: Add cycles to the cycle basis one by one. Check if the added cycle is a combination of some cycles already existing in the basis. If yes, delete this cycle. Step 4: Stop if the cycle basis has k cycles.

© The McGraw-Hill Companies, Inc., Detailed Steps for the Minimal Cycle Basis Problem Step 1 : A cycle basis corresponds to the fundamental set of cycles with respect to a spanning tree. a graph a spanning tree # of cycles in a cycle basis: = k = |E| - (|V|- 1) = |E| - |V| + 1 a fundamental set of cycles

© The McGraw-Hill Companies, Inc., Step 2: How to find all cycles in a graph? [Reingold, Nievergelt and Deo 1977] How many cycles are there in a graph in the worst case? In a complete digraph of n vertices and n(n-1) edges: Step 3: How to check if a cycle is a linear combination of some cycles? Using Gaussian elimination.

© The McGraw-Hill Companies, Inc., E.g.  2 cycles C 1 and C 2 are represented by a 0/1 matrix  Add C 3   on rows 1 and 3  on rows 2 and 3 : empty ∵ C 3 = C 1  C 2 Gaussian elimination

© The McGraw-Hill Companies, Inc., The 2-terminal One to Any Special Channel Routing Problem Def: Given two sets of terminals on the upper and lower rows, respectively, we have to connect each upper terminal to the lower row in a one-to-one fashion. This connection requires that the number of tracks used is minimized.

© The McGraw-Hill Companies, Inc., Two Feasible Solutions

© The McGraw-Hill Companies, Inc., Redrawing Solutions (a) Optimal solution (b) Another solution

© The McGraw-Hill Companies, Inc., At each point, the local density of the solution is the number of lines the vertical line intersects. The problem: to minimize the density. The density is a lower bound of the number of tracks. Upper row terminals: P 1,P 2, …, P n from left to right. Lower row terminals: Q 1,Q 2, …, Q m from left to right m > n. It would never have a crossing connection:

© The McGraw-Hill Companies, Inc., Suppose that we have the minimum density d of a problem instance. We can use the following greedy algorithm: Step 1 : P 1 is connected Q 1. Step 2 : After P i is connected to Q j, we check whether P i+1 can be connected to Q j+1. If the density is increased to d+1, try to connect P i+1 to Q j+2. Step 3 : Repeat Step2 until all P i ’ s are connected.

© The McGraw-Hill Companies, Inc., A Solution Produced by the Greedy Algorithm. d=1

© The McGraw-Hill Companies, Inc., The knapsack Problem n objects, each with a weight w i > 0 a profit p i > 0 capacity of knapsack: M Maximize Subject to 0  x i  1, 1  i  n

© The McGraw-Hill Companies, Inc., The knapsack problem is different from the 0/1 knapsack problem. In the 0/1 knapsack problem, x i is either 0 or 1 while in the knapsack problem, 0  x i  1.

© The McGraw-Hill Companies, Inc., The knapsack algorithm The greedy algorithm: Step 1: Sort p i /w i into nonincreasing order. Step 2: Put the objects into the knapsack according to the sorted sequence as far as possible. e.g. n = 3, M = 20, (p 1, p 2, p 3 ) = (25, 24, 15) (w 1, w 2, w 3 ) = (18, 15, 10) Sol: p 1 /w 1 = 25/18 = 1.32 p 2 /w 2 = 24/15 = 1.6 p 3 /w 3 = 15/10 = 1.5 Optimal solution: x 1 = 0, x 2 = 1, x 3 = 1/2